Final Exam Q6: Finding Problem with Torque Equation

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The discussion focuses on solving physics problems related to torque, average velocity, and motion equations. Key points include the importance of understanding the angle in torque calculations, specifically the angle between the radius vector and the force vector. Participants express confusion over average velocity calculations, emphasizing the need to consider displacement over time rather than simply averaging speeds. The conversation also touches on using kinematic equations to solve for time and velocity in various scenarios, highlighting the significance of horizontal and vertical components in motion. Overall, the thread serves as a collaborative effort to clarify complex physics concepts and problem-solving strategies.
  • #91
Hootenanny said:
I answered this above in post #86 :smile:

~H
that is a different problem
 
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  • #92
Alt+F4 said:
that is a different problem

Ahh, so it is. Perhaps it's time for me to take a break and get my fix of caffine :zzz:

~H
 
  • #93
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
That link isn't working.
 
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  • #94
sorry. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99
 
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  • #95
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114
Huh? The motion is parabolic, not straight line.

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
 
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  • #96
Doc Al said:
Huh? The motion is parabolic, not straight line.


(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
 
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  • #97
Alt+F4 said:
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
That's right. You know the coordinates of the landing point at time t.
 
  • #98
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?

and then the distance should equal 0 right/>
 
  • #99
Alt+F4 said:
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?
That will work.

and then the distance should equal 0 right?
When the ball lands, its coordinates are x = 110 m; y = 0.
 
  • #100
A satellite is in circular orbit at a fixed radius from the center of the Earth and with a constant speed. Which one of the following statements is correct about the satellite?

(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.


Ans:C


WHy so? Constant Speed = Constant Velocity
 
  • #101
Okay so for a problem, say i wana find how far must u bring ur hands across the string to get a frequency of 350. How would u go about solving that
 
  • #102
Alt+F4 said:
A satellite is in circular orbit at a fixed radius from the center of the Earth and with a constant speed. Which one of the following statements is correct about the satellite?

(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.


Ans:C


WHy so? Constant Speed = Constant Velocity
Careful! Velocity and acceleration are both vectors--direction counts.
 
  • #103
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 Question 18

(5)(6) + (2) ( 20) - (x)(25) = 0

X = 2.8 why is it 2
 
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  • #104
well

Px = 0
Py = 0
 
  • #105
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 Question 18

(5)(6) + (2) ( 20) - (x)(25) = 0

X = 2.8 why is it 2
First things first. Momentum is a vector. Start by finding the total momentum of the 5-kg and 2-kg pieces.
 
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  • #106
okay so Total Mom = (5)(6) for X, For Y it is (2)(20)

so for X = 30
so for Y = 40

so the vecotr of the third one

u get an angle of 53.13
 
  • #107
Alt+F4 said:
okay so Total Mom = (5)(6) for X, For Y it is (2)(20)

so for X = 30
so for Y = 40

so the vecotr of the third one

u get an angle of 53.13
So what's the magnitude of the total momentum (of those two pieces)?
 
  • #108
50 = MV
got it thanks
 
  • #109
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

Question 22

15*9.8*50 / 80 = 91.8 ~92

What exactly is the formula since i don't get why you do that, i just have it memorized Why multiply by radii?
 
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  • #110
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp04

Question 22

15*9.8*50 / 80 = 91.8 ~92

What exactly is the formula since i don't get why you do that, i just have it memorized Why multiply by radii?
For the cylinder to be in equilibrium, the torques must balance. Torque depends on the moment arm.
 
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  • #111
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?

So can u just explain to me When say Gravity is negative etc..


Does - Acceleration mean that ur slowing down? Deacc
 
  • #112
Alt+F4 said:
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?

So can u just explain to me When say Gravity is negative etc..


Does - Acceleration mean that ur slowing down? Deacc


Think about it this way. In general, g is only the magnitude of the acceleration in the y direction. Or, a_y=a_/smallfreefall=-g
Now, this is true when you choose the positive y-direction to point vertically upward. It is a_y that is negative, not g. Does that help?
 
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  • #113
Alt+F4 said:
Okay i need this straighen out before exam, An elevator going downward say it is Accelerating. Does that Mean Acc is - or is it postive?
Acceleration is the rate of change of velocity. It's a vector and has direction. Calling acceleration + or - is just a sign convention to indicate direction. Usually, + means up and - means down. Note that acceleration and velocity are not necessarily in the same direction. Just because the elevator is moving down, doesn't mean it's acceleration is downward. (It could be slowing down, for instance.)

So can u just explain to me When say Gravity is negative etc..
All that means is that the force of gravity (and the acceleration due to gravity) acts downward, the negative direction.

Does - Acceleration mean that ur slowing down?
Not at all. Acceleration just means that your velocity is changing. Toss a ball straight up into the air. The acceleration is always downward (negative). As the ball rises, it slows; as it falls, it speeds up. But it's always accelerating down.

"Deacceleration" is a confusing term; I would avoid it. (It sometimes is used to mean "slowing down", but it's more helpful to say that your acceleration is negative (compared to your velocity).)

Note that you can have acceleration without speeding up or slowing down. (Think circular motion.)
 

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